Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m;
int g[2][105][105];
vector<pair<string, int>> ans[2];
bool valid[2];
int main() {
  cin >> n >> m;
  int zero = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &g[0][i][j]);
      g[1][i][j] = g[0][i][j];
      if (!g[0][i][j]) zero++;
    }
  }
  if (zero == n * m) return 0 * printf("0");
  for (int i = 0; i < n; i++) {
    int low = 555;
    for (int j = 0; j < m; j++) {
      low = min(low, g[0][i][j]);
    }
    if (low) {
      for (int j = 0; j < m; j++) {
        g[0][i][j] -= low;
      }
      while (low--) {
        ans[0].emplace_back(string("row"), i + 1);
      }
    }
  }
  for (int i = 0; i < m; i++) {
    int low = 555;
    for (int j = 0; j < n; j++) {
      low = min(low, g[0][j][i]);
    }
    if (low) {
      for (int j = 0; j < n; j++) {
        g[0][j][i] -= low;
      }
      while (low--) {
        ans[0].emplace_back(string("col"), i + 1);
      }
    }
  }
  valid[0] = 1;
  for (int i = 0; i < n && valid[0]; i++)
    for (int j = 0; j < m && valid[0]; j++)
      if (g[0][i][j]) valid[0] = 0;
  for (int i = 0; i < m; i++) {
    int low = 555;
    for (int j = 0; j < n; j++) {
      low = min(low, g[1][j][i]);
    }
    if (low) {
      for (int j = 0; j < n; j++) {
        g[1][j][i] -= low;
      }
      while (low--) {
        ans[1].emplace_back(string("col"), i + 1);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    int low = 555;
    for (int j = 0; j < m; j++) {
      low = min(low, g[1][i][j]);
    }
    if (low) {
      for (int j = 0; j < m; j++) {
        g[1][i][j] -= low;
      }
      while (low--) {
        ans[1].emplace_back(string("row"), i + 1);
      }
    }
  }
  valid[1] = 1;
  for (int i = 0; i < n && valid[1]; i++)
    for (int j = 0; j < m && valid[1]; j++)
      if (g[1][i][j]) valid[1] = 0;
  if (!valid[0] && !valid[1]) return 0 * printf("-1");
  if (valid[0] && valid[1]) {
    if (ans[0].size() < ans[1].size()) {
      printf("%d\n", ans[0].size());
      for (auto const &e : ans[0]) {
        cout << e.first << " " << e.second << "\n";
      }
    } else {
      printf("%d\n", ans[1].size());
      for (auto const &e : ans[1]) {
        cout << e.first << " " << e.second << "\n";
      }
    }
  } else if (valid[0]) {
    printf("%d\n", ans[0].size());
    for (auto const &e : ans[0]) {
      cout << e.first << " " << e.second << "\n";
    }
  } else {
    printf("%d\n", ans[1].size());
    for (auto const &e : ans[1]) {
      cout << e.first << " " << e.second << "\n";
    }
  }
}